The actual force on the ground from a falling body's impact

[peter010]

1. problem:

a body of mass started to fall from rest, and from S elevation. what is the equation that calculate the force of what this body will hit the ground?

Homework Equations

:[/B]

F=mg // which is used to calculate the force of the body on the ground when it is already on the ground

The Attempt at a Solution

[/B]
F_new=F*S

// I am here taking S as a unitless number that would multiply the force over distance, but really don't know if this is the right handling of this problem.

May you help please/

 

[drvrm]

a body of mass m is threw, from rest, and from S elevation. what is the equation that calculate the force of what this body will hit the ground?

Homework Equations

:[/B]

F=mg // which is used to calculate the force of the body on the ground when it is already on the ground

The Attempt at a Solution

F_new=F*S

// I am here taking S as a unitless number that would multiply the force over distance, but really don't know if this is the right handling of this problem.

May you help please/[/B]

i could not get the meaning of your equation for Fnew
pl.try to answer the question on the concept of force -how one defines a force ...?..
is force related to mass or its velocity or both ?
what Newton's laws of motion say about force ?

 

[jbriggs444]

a body of mass m is threw, from rest, and from S elevation. what is the equation that calculate the force of what this body will hit the ground?

If you drop a rock into a puddle of thick mud, do you imagine that the impact force will be the same as if you drop it onto a steel plate?
If you drop a rock into a puddle of thick mud, do you imagine the impact will last for the same duration as if you drop it onto a steel plate?

Where did you find this question asked?

 

[peter010]

F_new: the foce of the body when hitting the ground after being fell

F=mg ...... (1)
F=m[v2-v1]/t
//v1=0
F=[m * v2]/t

Hi, i did applied Newton law -> v2^2 = 2gS

and i got finally, the rule: F=[ m * (2*g*S)^0.5] /t ...... (2)

lets assume: m=100 kg, g=9.81 m/s^2, s=0.3 meter, t=1sec,,

the confusion is:
- if i applied equation 1: i get the answer: 981 Newton (a) // represents as if the body at rest

- if i applied equation 2: i get the answer: 242.6 Newton, (b) // represents when body is just hitting the ground after being fell.

i get here the answer (b) less than the answer (a) !

Any correction ?>>>!

 

[peter010]

If you drop a rock into a puddle of thick mud, do you imagine that the impact force will be the same as if you drop it onto a steel plate?
If you drop a rock into a puddle of thick mud, do you imagine the impact will last for the same duration as if you drop it onto a steel plate?

Where did you find this question asked?

Absolutely not! this is wt I am talking about!

here i have same ground, but comparing between the body's force at rest VS a force of same body just hitting the ground after being fell...

 

[jbriggs444]

Absolutely not! this is wt I am talking about!

here i have same ground, but comparing between the body's force at rest VS a force of same body just hitting the ground after being fell...

OK, let's keep asking questions then...

Will the supporting force on a rock at rest in a mud puddle be the same as the supporting force on a rock at rest on a steel plate?

 

[peter010]

OK, let's keep asking questions then...

Will the supporting force on a rock at rest in a mud puddle be the same as the supporting force on a rock at rest on a steel plate?

if you meant by the supporting force is the reaction force, then it should be the same in both cases.

 

[peter010]

OK, let's keep asking questions then...

Will the supporting force on a rock at rest in a mud puddle be the same as the supporting force on a rock at rest on a steel plate?

oh! do you want to say, that its the same reacting force at final stage..

the steel plate will react on its surface, while the mud puddle will counteract the rock weight by making it accelerate inside it (as same as in my case)?

 

[jbriggs444]

I am trying to get you to think about whether the surface the object is falling onto changes the answer to your question.

 

[peter010]

I am trying to get you to think about whether the surface the object is falling onto changes the answer to your question.

in the mud puddle, clearly, the surface will get changed .. we will get a depth as an impact.

for steel surface, the effect will be less, and force will be discharged then as vibration.

 

[peter010]

but how this would answer me ) ?

 

[jbriggs444]

in the mud puddle, clearly, the surface will get changed .. we will get a depth as an impact.

for steel surface, the effect will be less, and force will be discharged then as vibration.

You asked a question: "What is the impact force?"

You appear to believe that the answer depends only upon height. I am suggesting that the answer will also depend on the surface that is impacted.

 

[peter010]

You asked a question: "What is the impact force?"

You appear to believe that the answer depends only upon height. I am suggesting that the answer will also depend on the surface that is impacted.

ok, I am sorry if i am not clear,,

any body at rest, like me now setting on my chair, my weight is then -> F=mg.

Now, let's say that i thew myself from the window at rest, and i fell over a car, i presume then, that my force just over the surface of this car is bigger than F=mg at rest, due to the distance I have fell though. So, i am asking what is the magnitude of that force when just hitting the surface of that car?but after our discussion, i found i should consider talking about the impact force, since effect of impact force is the way of how the body will discharge the energy earned by accelerating in space.

 

[jbriggs444]

Hi, i did applied Newton law -> v2^2 = 2gS

and i got finally, the rule: F=[ m * (2*g*S)^0.5] /t ...... (2)

lets assume: m=100 kg, g=9.81 m/s^2, s=0.3 meter, t=1sec,,

the confusion is:
- if i applied equation 1: i get the answer: 981 Newton (a) // represents as if the body at rest

- if i applied equation 2: i get the answer: 242.6 Newton, (b) // represents when body is just hitting the ground after being fell.

As I read the above:

You have a 100 kg object falling 0.3 meters in a 9.8 m/s² gravity field.
You assume without evidence that the ensuing impact will last for one second.
You correctly calculate the impact velocity using v² = 2gS.
You correctly divide the impact velocity by the impact duration to determine the impact acceleration and multiply by mass to get impact force.
You neglect the required upward force to resist gravity while the impact is taking place.
You correctly conclude that the net upward force required to produce the observed upward acceleration is 242.6 N.
You compare this to the gross upward supporting force required to hold the object in place against gravity.

 

[jbriggs444]

ok, I am sorry if i am not clear,,

any body at rest, like me now setting on my chair, my weight is then -> F=mg.

Now, let's say that i thew myself from the window at rest, and i fell over a car, i presume then, that my force just over the surface of this car is bigger than F=mg at rest, due to the distance I have fell though. So, i am asking what is the magnitude of that force when just hitting the surface of that car?but after our discussion, i found i should consider talking about the impact force, since effect of impact force is the way of how the body will discharge the energy earned by accelerating in space.

Discharging energy is not a commonly used phrase. Better wording would be in terms of how much work could be done by the impact. In physics, the amount of "work" can be calculated by force multiplied by distance. How much is the falling object deforming the ground upon which it lands and how much is the ground upon which it lands deforming the falling object?

Again, I strongly suggest that the ground upon which an object falls affects the distance covered during the impact, the time taken during the impact, the average force and average acceleration experienced during the impact, even if the impact velocity is the same.

 

[CWatters]

Peter... Have you studied the equations of motion under constant acceleration (eg SUVAT equations)?

If so perhaps try applying none of these to the deceleration phase that occurs when an object hits the ground...

Try assuming that it penetrates the ground a dist since "s" and use that as the stopping distance in the equation. What happens to the deceleration "a" if the ground is harder and hence the penetration and stopping distance is reduced? What happens to "a" if the stopping distance reduces to zero? What is the implication for the force?

 

[haruspex]

Peter, it might help you if you read https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/. Note in particular section 2.

 

[peter010]

Thanks indeed for you all )

 

[FactChecker]

The responses above are trying to get you to think of force in terms of F=mA. The force is directly determined by the acceleration (or deceleration). Will the deceleration at any time be the same if it falls into mud or water or steel? You need to work from there and you might need more information to calculate it. But the force is spread over time and changes, so that might get messy. Are you sure that you have the correct problem statement? I suggest that you keep this in mind and review the hints others have given above.